Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t^2 + t - 20}{3t + 15} \times \dfrac{-4t - 12}{t^2 - 4t} $
Explanation: First factor the quadratic. $x = \dfrac{(t - 4)(t + 5)}{3t + 15} \times \dfrac{-4t - 12}{t^2 - 4t} $ Then factor out any other terms. $x = \dfrac{(t - 4)(t + 5)}{3(t + 5)} \times \dfrac{-4(t + 3)}{t(t - 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (t - 4)(t + 5) \times -4(t + 3) } { 3(t + 5) \times t(t - 4) } $ $x = \dfrac{ -4(t - 4)(t + 5)(t + 3)}{ 3t(t + 5)(t - 4)} $ Notice that $(t + 5)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -4\cancel{(t - 4)}(t + 5)(t + 3)}{ 3t(t + 5)\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $x = \dfrac{ -4\cancel{(t - 4)}\cancel{(t + 5)}(t + 3)}{ 3t\cancel{(t + 5)}\cancel{(t - 4)}} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $x = \dfrac{-4(t + 3)}{3t} ; \space t \neq 4 ; \space t \neq -5 $